Factor: bca² + bcd² + adb² + adc²

[SweatingBear]

The task is to factor

\(\displaystyle bca^2 + bcd^2 + adb^2 +adc^2\),

which undeniably is a non-trivial one. It turns out that the expression can be factored into \(\displaystyle (ab + cd)(ac + bd)\). I have absolutely no idea what strategy to employ in non-trivial cases such as these (although I did attempt factoring by grouping, however to no real avail). Forum, got any clues/suggestions/tips? Anything helps!

 

[kaliprasad]

The task is to factor

\(\displaystyle bca^2 + bcd^2 + adb^2 +adc^2\),

which undeniably is a non-trivial one. It turns out that the expression can be factored into \(\displaystyle (ab + cd)(ac + bd)\). I have absolutely no idea what strategy to employ in non-trivial cases such as these (although I did attempt factoring by grouping, however to no real avail). Forum, got any clues/suggestions/tips? Anything helps!

try to group it
bca^2 + bcd^2 + adb^2 +adc^2
= bc(a^2+d^2) + ad(b^2 + c^2)
no

bca^2 + adb^2 + bcd^2 + adc^2
= ab( ac+ bd) + bc(bd + ad) ( I got it)
= ( ac+bd)(ab+ cd)

it is done

did you miss something

 

[SuperSonic4]

The task is to factor

\(\displaystyle bca^2 + bcd^2 + adb^2 +adc^2\),

which undeniably is a non-trivial one. It turns out that the expression can be factored into \(\displaystyle (ab + cd)(ac + bd)\). I have absolutely no idea what strategy to employ in non-trivial cases such as these (although I did attempt factoring by grouping, however to no real avail). Forum, got any clues/suggestions/tips? Anything helps!

Essentially it comes down to trial and error and practice.

I would seek to factor \(\displaystyle bc \) from the first two terms and \(\displaystyle ad\) from the second two simply because they stand out to me.

\(\displaystyle bc(a^2+d^2) + ad(b^2+c^2)\). This just leads to a dead end though.

---

Let's try \(\displaystyle ab\) and \(\displaystyle cd\) respectively (some question setters aren't very imaginative (Wink))

\(\displaystyle ab(ac +db) + cd(bd + ac) = ab(ac+bd) + cd(ac+bd)\)

Now I have \(\displaystyle ac+bd\) in both terms I can factor again to get the answer:

\(\displaystyle (ac+bd)(ab+cd)\)

 

[SweatingBear]

Thanks a lot, looks like trial-and-error will have to do.

 

[CellCoree]

I would suggest approaching this problem using algebraic techniques and properties. One strategy could be to look for common factors among the terms, such as b or d. We can rewrite the expression as (bca^2 + bcd^2) + (adb^2 + adc^2) and then factor out the common factor of b, giving us b(ca^2 + cd^2) + d(ad^2 + ac^2). From there, we can factor out another common factor of a from the first grouping and c from the second grouping, giving us ab(c + d) + cd(a + d). We can then use the distributive property to rearrange the terms and factor out the common factor of (c + d), giving us (ab + cd)(c + d). This is close to the desired factorization of (ab + cd)(ac + bd), and we can rearrange the terms using the commutative property to get the final result. It is important to remember that factoring can often involve trial and error, and it may take multiple attempts to find the correct solution. Additionally, it can be helpful to check the final factorization by expanding it back out to the original expression to ensure it is correct.